Two numbers add to 6. Let U denote the sum of the first number and twice
the square of the second number. What is the smallest
possible value for U?

Let's denote the two numbers as \( x \) and \( y \). We are given:

\[
x + y = 6
\]

We want to minimize \( U \), which is defined as:

\[
U = x + 2y^2
\]

Since \( x + y = 6 \), we can express \( x \) in terms of \( y \):

\[
x = 6 - y
\]

Substitute this into the expression for \( U \):

\[
U = (6 - y) + 2y^2
\]

Simplifying, we get:

\[
U = 6 - y + 2y^2
\]

Now, to find the minimum value of \( U \), we'll take the derivative of \( U \) with respect to \( y \) and set it equal to zero:

\[
\frac{dU}{dy} = -1 + 4y
\]

Setting the derivative equal to zero:

\[
-1 + 4y = 0
\]

Solving for \( y \):

\[
4y = 1 \quad \Rightarrow \quad y = \frac{1}{4}
\]

Next, substitute \( y = \frac{1}{4} \) back into the expression for \( U \) to find the minimum value:

\[
U = 6 - \frac{1}{4} + 2\left(\frac{1}{4}\right)^2
\]

Simplify this expression:

\[
U = 6 - \frac{1}{4} + 2 \times \frac{1}{16}
\]

\[
U = 6 - \frac{1}{4} + \frac{1}{8}
\]

\[
U = 6 - \frac{2}{8} + \frac{1}{8} = 6 - \frac{1}{8} = \frac{48}{8} - \frac{1}{8} = \frac{47}{8} = 5.875
\]

So, the smallest possible value for \( U \) is \( \boxed{5.875} \).

Would you like any more details or have any questions?

Here are some related questions:

1. What is the significance of taking the derivative in this problem?
2. How would you solve this problem if the relationship between \( x \) and \( y \) were different?
3. What happens to the minimum value of \( U \) if the coefficient of \( y^2 \) changes?
4. Can \( U \) have a maximum value under the given constraints?
5. How can you use the second derivative to verify that the value we found is a minimum?

**Tip:** Always check your derivative calculations, especially when finding minimum or maximum values in optimization problems.

Explore how to minimize U in equations involving algebraic optimization and derivatives. Learn step-by-step with detailed explanations and examples. Suitable for advanced high school students and college-level learners.

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